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Question Number 147144 Answers: 1 Comments: 0

prove that equation of a circle passing through the points of intersection of a circle S=0 and a line L=0 can be taken as S+λL=0 where λ is a parameter

$${prove}\:{that}\: \\ $$$${equation}\:{of}\:{a}\:{circle}\:{passing}\:{through} \\ $$$${the}\:{points}\:{of}\:{intersection}\:{of}\:{a}\:{circle} \\ $$$${S}=\mathrm{0}\:{and}\:{a}\:{line}\:{L}=\mathrm{0}\:{can}\:{be}\:{taken}\:{as} \\ $$$${S}+\lambda{L}=\mathrm{0}\:{where}\:\lambda\:{is}\:{a}\:{parameter} \\ $$

Question Number 147137 Answers: 1 Comments: 0

Question Number 147135 Answers: 3 Comments: 0

Question Number 147134 Answers: 0 Comments: 2

(√(2sin(x)−1)) ∙ sin(x) > 0

$$\sqrt{\mathrm{2}{sin}\left({x}\right)−\mathrm{1}}\:\centerdot\:{sin}\left({x}\right)\:>\:\mathrm{0} \\ $$

Question Number 147130 Answers: 1 Comments: 2

Question Number 147115 Answers: 0 Comments: 0

pleaae there is challenge to this question as to whether the answer is ((43)/6) OR −((1187)/(42)) please help Question simplify 37(1/2) ÷ (5/9) of ((4/7)+(1/5))−80(1/3). the same question but different answer from different books

$$\mathrm{pleaae}\:\mathrm{there}\:\mathrm{is}\:\mathrm{challenge}\:\mathrm{to}\:\mathrm{this}\: \\ $$$$\mathrm{question}\:\mathrm{as}\:\mathrm{to}\:\mathrm{whether}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is} \\ $$$$\:\:\:\:\:\frac{\mathrm{43}}{\mathrm{6}}\:\:\:\mathrm{OR}\:\:\:−\frac{\mathrm{1187}}{\mathrm{42}}\:\:\mathrm{please}\:\mathrm{help} \\ $$$$\mathrm{Question}\: \\ $$$$\mathrm{simplify}\:\:\mathrm{37}\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{\div}\:\frac{\mathrm{5}}{\mathrm{9}}\:\mathrm{of}\:\left(\frac{\mathrm{4}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{80}\frac{\mathrm{1}}{\mathrm{3}}. \\ $$$$\:\mathrm{the}\:\mathrm{same}\:\mathrm{question}\:\mathrm{but}\:\mathrm{different}\: \\ $$$$\mathrm{answer}\:\mathrm{from}\:\mathrm{different}\:\mathrm{books} \\ $$

Question Number 147122 Answers: 1 Comments: 0

lim_(n→∞) Σ_(k=1) ^n 2^k ∙((2)^(1/2^k ) −1)^2 = ?

$$\underset{\boldsymbol{{n}}\rightarrow\infty} {{lim}}\:\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\mathrm{2}^{\boldsymbol{{k}}} \centerdot\left(\sqrt[{\mathrm{2}^{\boldsymbol{{k}}} }]{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \:=\:?\: \\ $$

Question Number 147344 Answers: 1 Comments: 0

lim_(n→∞) ∫_( 0) ^( 1) log (((1+sin^n x)/(1+x+x^n )))dx = ?

$$\underset{{n}\rightarrow\infty} {{lim}}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:{log}\:\left(\frac{\mathrm{1}+{sin}^{\boldsymbol{{n}}} \boldsymbol{{x}}}{\mathrm{1}+{x}+{x}^{\boldsymbol{{n}}} }\right){dx}\:=\:? \\ $$

Question Number 147119 Answers: 2 Comments: 0

if x>−1 , q≥2 then: (1+x)^q ≥ 1+qx+(q−1)x^2

$${if}\:\:\:{x}>−\mathrm{1}\:,\:{q}\geqslant\mathrm{2}\:\:\:{then}: \\ $$$$\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}} \:\geqslant\:\mathrm{1}+{qx}+\left({q}−\mathrm{1}\right){x}^{\mathrm{2}} \\ $$

Question Number 147118 Answers: 1 Comments: 0

Question Number 147116 Answers: 2 Comments: 5

a+b+c=1 a^2 +b^2 +c^2 =1 a^3 +b^3 +c^3 =1 a , b , c =?

$${a}+{b}+{c}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{1} \\ $$$${a}\:,\:{b}\:,\:{c}\:=? \\ $$

Question Number 147108 Answers: 2 Comments: 0

Question Number 147103 Answers: 2 Comments: 0

Question Number 147101 Answers: 1 Comments: 0

find U_n =∫_0 ^1 (1+x^2 )(1+x^4 )....(1+x^2^n )dx

$$\mathrm{find}\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)....\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}^{\mathrm{n}} } \right)\mathrm{dx} \\ $$

Question Number 147100 Answers: 0 Comments: 0

findA_n = ∫_0 ^1 x(x+1)(x+2)....(x+n)dx

$$\mathrm{findA}_{\mathrm{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)....\left(\mathrm{x}+\mathrm{n}\right)\mathrm{dx} \\ $$

Question Number 147093 Answers: 3 Comments: 2

(1)lim_(x→π/2) ((cos 4x−cos 2x−2)/((2x−π)^2 )) =? (2)lim_(x→0) ((sin 3x+sin 6x−sin 9x)/x^3 ) =? (3)lim_(x→π/4) ((sec^2 x−2tan x)/((x−(π/4))^2 )) =? (4)lim_(x→0) ((12−6x^2 −12cos x)/x^4 )=? (5)lim_(x→0) ((sin^2 x−sin^2 2x+3x^2 )/x^4 )=?

$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{2}{x}−\mathrm{2}}{\left(\mathrm{2}{x}−\pi\right)^{\mathrm{2}} }\:=? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{3}{x}+\mathrm{sin}\:\mathrm{6}{x}−\mathrm{sin}\:\mathrm{9}{x}}{{x}^{\mathrm{3}} }\:=? \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{2tan}\:{x}}{\left({x}−\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} }\:=? \\ $$$$\left(\mathrm{4}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{12}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{12cos}\:{x}}{{x}^{\mathrm{4}} }=? \\ $$$$\left(\mathrm{5}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} }=? \\ $$

Question Number 147091 Answers: 1 Comments: 0